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Diagonalisation and matrix powers

A = PDP⁻¹, computing Aⁿ, and the matrices that cannot be diagonalised.

With a full set of eigenvectors, A = PDP⁻¹ where D is diagonal, so Aⁿ = PDⁿP⁻¹: Fibonacci in closed form, Markov chains at equilibrium. Picture it: the transformation stretching along eigenvector directions and nothing else. Think it: symmetric matrices always diagonalise with orthogonal eigenvectors (the spectral theorem); defective ones need Jordan form.

कार्यरत उदाहरण · eigenvalues of [[4,1],[2,3]]

Linear-Algebra · Eigenvalues

Eigenvalues of [[4,1],[2,3]]

\left[\begin{matrix}4 & 1\\2 & 3\end{matrix}\right]

चरणानुक्रमे

  1. \det(A - \lambda I) = 0

    Eigenvalues are the roots of the characteristic polynomial.

  2. \det\left[\begin{matrix}4 - \lambda & 1\\2 & 3 - \lambda\end{matrix}\right] = 0

    Subtract λ from the diagonal.

  3. \lambda^{2} - 7 \lambda + 10 = 0

    Expand the determinant.

  4. \left(\lambda - 5\right) \left(\lambda - 2\right) = 0

    Factor.

  5. \lambda = 5, \lambda = 2

    Eigenvalues (with multiplicity).

  6. \lambda = 2:\ (A - \lambda I)\mathbf{v} = 0 \Rightarrow \mathbf{v} = \left[\begin{matrix}- \frac{1}{2}\\1\end{matrix}\right]

    Solve (A − 2I)v = 0 for a basis eigenvector.

  7. \lambda = 5:\ (A - \lambda I)\mathbf{v} = 0 \Rightarrow \mathbf{v} = \left[\begin{matrix}1\\1\end{matrix}\right]

    Solve (A − 5I)v = 0 for a basis eigenvector.

उत्तर उघडा
\lambda = 5,\; \lambda = 2

आता तुम्ही प्रश्न निवडा, किंवा स्वतःचा प्रश्न टाइप करा किंवा रेखाटून दाखवा. प्रत्येक टप्प्यात, चित्रे आहेत, तुम्ही विचारण्याआधी उत्तर लपवलेले आहे.

स्वतःचे कार्य चालू ठेवा

मोफत खाते प्रत्येक शिकवणी वर टिपण्णी जोडते, तुम्ही पूर्ण केलेल्या गोष्टीचा नोंदवही, तुमच्या सोडविलेल्या समस्या एका ठिकाणी, आणि या पानाच्या विषयी तुम्ही विचारू शकता अशा शिक्षकाला. गणित स्वतःच सर्वांसाठी खुले आहे, दाखलन झालेले असो किंवा नसो.

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कसे कराल: Diagonalisation and matrix powers

  1. Eigenvalues are the roots of the characteristic polynomial.
  2. Subtract λ from the diagonal.
  3. Expand the determinant.
  4. Factor.
  5. Eigenvalues (with multiplicity).
  6. Solve (A − 2I)v = 0 for a basis eigenvector.
  7. Solve (A − 5I)v = 0 for a basis eigenvector.

लोक विचारतात ते प्रश्न

What does a determinant mean geometrically?

It is the factor by which the matrix scales area (2×2) or volume (3×3), with a negative sign if orientation flips. Zero means the matrix flattens space and cannot be undone.

What is an eigenvector?

A direction the matrix does not turn: it only stretches it by the eigenvalue. Along eigenvectors a complicated matrix acts like multiplication by a number.

Why is matrix multiplication not commutative?

Because a matrix is a transformation and AB means "do B, then A". Rotating then reflecting is not the same as reflecting then rotating.

अधिक माहिती Linear Algebra

DeterminantsMatrix inverseEigenvalues and eigenvectorsMatrix multiplicationRow reductionVector spaces, span and linear independenceOrthogonality, projections and least squaresLinear transformations and change of basis